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A 100 mL solution of Na3PO4 contains 3.45 g of sodium. The molarity of the solution is _____ x 10⁻² mol L⁻¹ (nearest integer). [Atomic masses: Na = 23.0 u, O = 16.0 u, P = 31.0 u]

1. 50
2. 15
3. 5
4. 150

Correct answer: 50

Solution

Moles of Na = 3.45/23 = 0.15; each Na3PO4 has 3 Na so moles of salt = 0.05; molarity = 0.05/0.1 = 0.5 M = 50 x 10⁻² M.

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